Answer:
Step-by-step explanation:
Using the alternative hypothesis (µ < µ0),
To find the p-value with test statistic -1.25 and assuming a standard level of significance of 0.05, using a p value calculator, the p-value is 0.1057 which is great that 0.05. Thus, the results is not significant.
Using the p value calculation.
1. Check the left tailed z table as the test statistic is negative,
2. Then find the probabilitythat z is greater than your test statistic (look up your test statistic on the z-table- the value under 1.2 and 0.05 which is 0.8944
3. Then, find its corresponding probability, and subtract it from 1 to get your p-value- 1-0.8944 = 0.1056.
Let be y =f(x)= x3, f(-x) = (-x)^3 = - x^3 = - f(-x), f(-x) = (-x)^3, f(-x) = - f(<span>-x)
the equation is odd</span>
4 + 3 * (7 - 2)
4 + 3 * 5
4 + 15
19 <==
Are you solving for x and y? 5y + 2(6x) would just be 5y + 12x
Multiply by two
2/5 x 2/1 = 4/5
